HDU3530-Subsequence

Subsequence

                                                                         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                    Total Submission(s): 6539    Accepted Submission(s): 2184

Problem Description

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

 

Input

There are multiple test cases.
For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000].
Proceed to the end of file.

 

Output

For each test case, print the length of the subsequence on a single line.

 

Sample Input

5 0 01 1 1 1 15 0 31 2 3 4 5

 

Sample Output

54

 

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

 

Recommend

zhengfeng

 

题意：求一个最长的最大和最小的差在[m, k]范围的序列

解题思路：维护一个最大序列和一个最小序列，然后判断是否符合范围就行了，注意题目并没有说明m和k之间的关系

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;const int MAXN=100010;int q1[MAXN],q2[MAXN];int rear1,head1;int rear2,head2;int a[MAXN];int main(){    int n,m,k;    while(~scanf("%d %d %d",&n,&m,&k))    {        rear1=head1=0;        rear2=head2=0;        int ans=0;        int now=1;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            while(head1<rear1&&a[q1[rear1-1]]<a[i]) rear1--;            while(head2<rear2&&a[q2[rear2-1]]>a[i]) rear2--;            q1[rear1++]=i;            q2[rear2++]=i;            while(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>k)            {                if(q1[head1]<q2[head2]) now=q1[head1++]+1;                else now=q2[head2++]+1;            }            if(head1<rear1&&head2<rear2&&a[q1[head1]]-a[q2[head2]]>=m)                ans=max(ans,i-now+1);        }        printf("%d\n",ans);    }    return 0;}